I'm reading an example for abstract algebra and I would like to clarify some terms. The example is showing the number of homomorphisms from $\mathbb Z $ to $S_3$.

Take a homomorphism $f$ from $\mathbb Z$ to $S_3$. Since $1$ is a generator of $\mathbb Z$, $f$ is completely determined by the image of $1$. Since $1$ has infinite order, there are no restrictions on the image of $1$. Thus each element of $S_3$ is a possible image for $1$. There are $6$ homomorphisms from $\mathbb Z$ to $S_3$.

What does "completely determined by the image of $1$" mean? I thought about it a little bit, and it seems that since $1$ is a generator of $\mathbb Z$, any element in $\mathbb Z$ can be generated from $1^k$. When it says image of $1$, does it mean $f(1^k)$, where $k$ is an integer? In general, what does the "image" of an element under homomorphism mean? I thought for an element $g$ in group $G$, and for a homomorphism $h$, the image of $g$ under $h$ is $h(g)$. How can that determine the entire homomorphism?

Thanks for your help.

  • $\begingroup$ As $\mathbb Z$ is written additively, it qwould be less confusing to write $k\cdot 1$ instead of $1^k$ (after all, it is supposed to mean $\underbrace{1+1+\ldots+1}_{k}$ $\endgroup$ – Hagen von Eitzen Oct 19 '14 at 20:32
  • $\begingroup$ Thanks Hagen. How did you fix up the formatting? Would like to learn for future posts. $\endgroup$ – jstnchng Oct 19 '14 at 20:52
  • $\begingroup$ @JC1397 You can use LaTeX both inline with $ and as a paragraph with $$ delimiters. Read the editing help for details. $\endgroup$ – Lukas Oct 19 '14 at 21:42

You are right. The image of $g\in A$ under a homomorphism $h\colon A\to B$ is the element $h(g)\in B$ to which $g$ is mapped.

That $f\colon \mathbb Z\to S_3$ is completely determined by the image of $1$ means that if you have two homomorphisms $f_,g\colon \mathbb Z\to S_3$ and the image of $1$ is the same for both (i.e. $f(1)=g(1)$) then the homomorphisms are the same (i.e. $f(x)=g(x)$ for all $x\in\mathbb Z$). More generally, if $A$ is a cyclic group with generator $a$ and $B$ is any group, then a homomorphism $f\colon A\to B$ is uniquely determined by the image of $a$. This follows from the homomorphism properties of $f$ and $g$ because $f(a)=g(a)$ gives us $f(a^{-1})=g(a^{-1})$ and then by induction $f(a^{\pm n})=g(a^{\pm n})$. Since $a$ generates $A$, this covers all elements of $A$. Even more generally, the corresponding statement holds also for groups with more generators.


It's a general property of algebraic structures, but let's limit ourselves to groups.

For any subset $S$ of $G$ there is the least subgroup $\langle S\rangle$ of $G$ containing $S$. It can be described as the set of all elements of the form $$ s_1^{\varepsilon_1}s_2^{\varepsilon_2}\dots s_n^{\varepsilon_n} $$ where $n$ is an arbitrary integer, $s_1,s_2,\dots,s_n\in S$ and $\varepsilon_k=\pm1$ for $k=1,2,\dots,n$.

The subset $S$ is said to be a generating set for $G$ if $\langle S\rangle=G$.

If $f\colon G\to G'$ and $g\colon G\to G'$ are homomorphisms, $S$ is a generating set for $G$ and $f(s)=g(s)$ for all $s\in S$, then $f=g$.

Indeed, the condition on $S$ tells us that, for each $x\in G$, there are $s_1,s_2,\dots,s_n\in S$ and $\varepsilon_k=\pm1$ for $k=1,2,\dots,n$ with $$ x=s_1^{\varepsilon_1}s_2^{\varepsilon_2}\dots s_n^{\varepsilon_n} $$ and so \begin{align} f(x)&=f(s_1^{\varepsilon_1}s_2^{\varepsilon_2}\dots s_n^{\varepsilon_n})\\ &=f(s_1)^{\varepsilon_1}f(s_2)^{\varepsilon_2}\dots f(s_n)^{\varepsilon_n}\\ &=g(s_1)^{\varepsilon_1}g(s_2)^{\varepsilon_2}\dots g(s_n)^{\varepsilon_n}\\ &=g(s_1^{\varepsilon_1}s_2^{\varepsilon_2}\dots s_n^{\varepsilon_n})\\ &=g(x) \end{align}

The notation on $\mathbb{Z}$ is additive, but it's just a question of notation. It's clear that $\{1\}$ is a generating set for $\mathbb{Z}$. Thus two homomorphisms $f\colon\mathbb{Z}\to S_3$ and $g\colon\mathbb{Z}\to S_3$ such that $f(1)=g(1)$ are the same homomorphism.

The group $\mathbb{Z}$ has however a very special property: if $G$ is a group, for any $x\in G$ there is a homomorphism $\varphi_x\colon\mathbb{Z}\to G$ with $\varphi_x(1)=x$. This is a stronger property than $\{1\}$ being a generating set (it's usually expressed by saying that $\mathbb{Z}$ is a free group on $\{1\}$).

So the number of homomorphisms from $\mathbb{Z}$ to $G$ is always the same as the cardinality of $G$.

  • $\begingroup$ I see. egreg, could you also explain why this is true for the group Z, and if this is true only for the group Z? Thanks for your answer. $\endgroup$ – jstnchng Oct 19 '14 at 21:00
  • $\begingroup$ @JC1397 Once you have selected $x\in G$, you can (and actually must) define $\varphi_x(n)=x^n$, for $n\in\mathbb{Z}$, because $n=\underbrace{1+1+\dots+1}_{n \text{ times}}$ for $n>0$, so necessarily $\varphi_x(n)=\underbrace{xx\dots x}_{n\text{ times}}$. Complete for $n<0$. $\endgroup$ – egreg Oct 19 '14 at 21:08
  • $\begingroup$ I see. Thanks a lot for a detailed answer. $\endgroup$ – jstnchng Oct 19 '14 at 21:33

It means that if we give the value of $f(1)$ then we have the value of $f(n)$ for all $n\in\Bbb Z$ since

$$f(n)=f(\underbrace{1+\cdots+1}_{n\;\text{times}})=f(1)\circ\cdots \circ f(1)$$

but since there's $6$ elements of $S_3$ then there's $6$ homomorphisms from $\Bbb Z$ to $S_3$.


Hint: use the property that for any homomorphism $\phi$, $\phi(xy) = \phi(x)\phi(y)$. Since $1$ generates $\mathbb{Z}$ under addition, i.e. for any positive integer $n$ we have $$n = \underbrace{1+1+\cdots+1}_{n \text{ times}}$$,

we have $$\phi(n) = \phi(\underbrace{1+1+\cdots+1}_{n \text{ times}}) = \underbrace{\phi(1) \circ \phi(1) \circ \cdots \circ \phi(1)}_{n \text{ times}}$$

Hence, as you can see, the image of $\mathbb{Z}$ under $\phi$ (that is, the set $\{\phi(z) \mid z \in \mathbb{Z}\}$ is completely determined by where $\phi$ takes $1$. Once we know what $\phi(1)$ is, we know that $\phi(n) = \phi(1)^{n}$ (or $n\phi(1)$, if you prefer additive notation), since $\phi$ is a homomorphism - there is no flexibility here.


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